# Partial-tree procedure

The following procedure `list->tree` converts an ordered list to a balanced binary tree. The helper procedure `partial-tree` takes as arguments an integer `n` and list of at least `n` elements and constructs a balanced tree containing the first `n` elements of the list. The result returned by `partial-tree` is a pair (formed with `cons` ) whose `car` is the constructed tree and whose `cdr` is the list of elements not included in the tree.

``````(define (list->tree elements)
(car (partial-tree elements (length elements))))

(define (partial-tree elts n)
(if (= n 0)
(cons '() elts)
(let ((left-size (quotient (- n 1) 2)))
(let ((left-result (partial-tree elts left-size)))
(let ((left-tree (car left-result))
(non-left-elts (cdr left-result))
(right-size (- n (+ left-size 1))))
(let ((this-entry (car non-left-elts))
(right-result (partial-tree (cdr non-left-elts)
right-size)))
(let ((right-tree (car right-result))
(remaining-elts (cdr right-result)))
(cons (make-tree this-entry left-tree right-tree)
remaining-elts))))))))
``````

a. Write a short paragraph explaining as clearly as you can how `partial-tree` works. Draw the tree produced by `list->tree` for the list `(1 3 5 7 9 11)`

b. What is the order of growth in the number of steps required by `list->tree` to convert a list of `n` elements?

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